electronic power conversion - tu delft ocwswitch mode dc-ac converters 7 spectrum ˆ ˆ control a...
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1 Challenge the future 1 Challenge the future
Electronic Power Conversion Switch mode DC-AC converters
2 Switch mode DC-AC converters
8. Switch mode DC-AC converters • Applications:
• AC motor drives • Uninterruptible Power Supplies (UPS)
• Categories of voltage-source inverters (VSI,VSC): • PWM inverters • Square-wave inverters • Single-phase inverters with voltage cancellation • Not considered: CSI (Current-Source Inverters, CSI)
Uni-directional power Bidirectional power (regenerative)
Source: Wikimedia Commons CC-BY-SA
3 Switch mode DC-AC converters
Switch-mode inverters – basics • Requirement for instantaneous power flow:
• Four-quadrant operation
4 Switch mode DC-AC converters
Phase leg (phase arm or NL: fase tak)
• All dc-ac topologies in Chapt. 8 based on phase arm • Point 0 is taken as reference for voltage, mostly not physically
available
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• Consider one leg of an H-bridge and vary vcontrol slowly
6 Switch mode DC-AC converters
Synthesis of a Sinusoidal Output by PWM
control triv v> 0A d1v = V2
control triv < v 0A d1v = V2
−
TA+ is on
TA- is on
è
è
7 Switch mode DC-AC converters
Spectrum
ˆˆ
controla
tri
Vm = V
1
sf
fm = f
Amplitude modulation ratio:
Frequency modulation ratio:
fs - switching frequency, carrier frequency f1 - modulating frequency, fundamental frequency
8 Switch mode DC-AC converters
Details of a Switching Time Period
Control voltage can be assumed constant during a switching time-period
0ˆ
ˆcontrol d
A control tritri
v VV = with v < V2V
1ˆ sincontrol controlv = V tω
0,1 1 1
ˆsin sinˆ
control d dA a
tri
V V VV = t = m t 2 2V
ω ω
’Moving average’ of vA0 :
where:
From (1) and (2):
(1)
(2)
9 Switch mode DC-AC converters
Modulation frequency
• Small mf (mf< 21) • Apply synchronous PWM to avoid subharmonics • Use methods to eliminate specific low order harmonics
• Large mf (mf> 21) • Good reproduction of reference wave • Side band harmonics are small • Switching harmonics can be eliminated with small filter
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Overmodulation
ˆ ˆ1,a control trim V V> >
Voltage spectrum:
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• Output voltage Fundamental as a Function of ma ;
• Shows the linear and the over-modulation region;
• square-wave operation in the limit
( )0 1ˆd dA
V V4 < V < 2 2π
Overmodulation:
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Square wave operation
( ) ( )0 10
ˆˆ A
A h
VV
h=( )0 1,max
ˆ d dA
V V4V = = 1.273 2 2π
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Single-phase inverters (half bridge) • Current sharing between capacitors; • i0 cannot have a dc-component è no transformer saturation
problem; • Switch utilisation:
• VT = Vd • IT = Io
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Full-bridge converter
0 0 0 0( ) ( ) ( ) 2 ( )A B Av t = v t v t v t− =
0,1ˆ
a dV = m V
,ˆd d0 14V < < V V π
(ma<1)
(ma>1)
PWM with Bipolar voltage switching
Preferred to half-bridge at high power levels
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DC-side current (* = “averaged” currents) fs è ∞ , so L and C è 0
*0 0
0 1 0 1
0 0 0 0 1
( ) ( ) ( )
sin sin( )cos cos( )
d dV i t = v t i t
= 2 V t 2 I t - V I - V I 2 t -
ω ω φφ ω φ
⋅=
* 0 0 0 01( ) cos cos( )d
d d
V I V Ii t - 2 t -V V
φ ω φ=
From power balance:
Input current: • DC component • LF component of double fundamental frequency • Strong HF components (steep slopes)
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Full-bridge converter – PWM with unipolar voltage switching
• Double control voltage • Same output voltage and
magnitude as bipolar switching • Lower ripple current
(about 50%) • Less harmonics
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Square wave operation – output voltage control
0 0ˆ( ) cos( )
sin( )
2h
2
d
2V = v h d
4= V hh
π
π θ θπ
βπ
−∫
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Voltage and current ripple
0 01( ) ( ) ( )ripplev t v t v t= +
0 01( ) ( ) ( )ripplei t i t i t= +
01 01 1 01j Lω= +V E I
0( ) ( )
t
ripple ripple1i t = v d kL
ζ ζ +∫
Apply superposition:
Fundamental (phasors):
Ripple current:
(b): largest ripple during zero crossing of voltage
Square-wave PWM bipolar
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Switch utilisation ratio (SUR)
TT
total apparent converter power SSUR = q V I⋅
0
Tswitches
PSUR P
=∑
01,max ,maxd4V = V
2π 0I
,maxT dV = V 0,maxTI = 2 I
,max 0,max01,max 0,max0
,max 0,max ,max 0,max
d
T d d
4 V IV IP 12SUR = = q P 2q V 2 I 4 V 2 I
ππ
⋅⋅= =
⋅ ⋅
Example: Full bridge square-wave with sinusoidal current and output current I0
Output: ;
Switch T: ;
q = total number of switches
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Effect of blanking time
• Blanking time (dead time) tΔ to avoid shoot-through
( )
( )
d As
AN
d As
t + V i > 0T
V = t - V i 0T
Δ
Δ
Δ<
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Effect of blanking time (2)
AB AN BNV = V VΔ Δ − Δ
( )
( )
0ds
AB
0ds
2 t + V > 0iTV =
2 t - V < 0iT
Δ
Δ
Δ
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Voltage versus Current Source • 1) VSI with unipolar or bipolar modulation
• Behaves as voltage source • 2a) Current mode modulation:
• VSI with hysteresis modulation • Behaves as current source • Variable switching frequency • Constant current ripple amplitude
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Voltage versus Current Source (2)
• 2b) VSI with Fixed frequency current control (triangular modulation with feedback)
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Rectifier mode operation • Phasor representation for fundamental:
• The phase and amplitude of vAn(t) can be set through the input signal vcontrol of the modulator
An A L = +V E V L ss = j LωV I An AA
s
= j Lω−V EIwhere
sL s
div = L dt
sL s
div = L dt
25 Switch mode DC-AC converters
Figure 2
Figure 2 shows a single-‐phase inverter connected to a single phase induc6on motor with counter emf e0. The output voltage v0 of the inverter is obtained by bipolar voltage switching modula6on scheme. To obtain a low distor6on linear modula6on is applied (no overmodula6on; ma<1).
Given is further: Vd = 350V (DC link voltage) ω1 (fundamental frequency of vo and e0) ω1,nom =2π50 rad/s (nominal value of ω1) V01,nom =230 V (nominal rms value of fundamental of v0) e0 (counter emf) L = 30 mH (inductance of machine) fs=7.5 kHz (switching frequency) At nominal speed and nominal voltage the input power of the loaded drive is 2kW at cos φ1=0.8.
(a) Draw the relevant circuit 6me diagrams to show the opera6on of the inverter. Indicate which switch is on at what 6me period. Sketch equivalent circuit models to calculate the fundamental component of the current io and the ripple component of the current io. Calculate the maximum value of the peak-‐to-‐peak current ripple in io that is caused by the switching. (b) Calculate the rms value of the fundamental of io when the machine runs at rated speed and rated power. Sketch a phasor diagram with the phasors of e0, v0 and i0. (c) Calculate the modula6on ra6o ma such that the machine runs at nominal speed and nominal voltage. (d) What will change in the answers to problems b) and c) if unipolar switching is used instead of bipolar switching? (e) Calculate the low-‐frequency (<1 kHz) peak-‐to-‐peak voltage ripple ∆Vd assuming that the current i1 is constant and C=1mF.
i1
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Given is a single-‐phase full bridge inverter opera6ng in a square-‐wave mode. The dc-‐voltage is 244V and the frequency of the output voltage that supplies a motor load is 47Hz. The inductance is L = 100 mH. Calculate the peak value of the ripple in the output current.
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Image credits
• All uncredited diagrams are from the book “Power Electronics: Converters, Applications, and Design” by N. Mohan, T.M. Undeland and W.P. Robbins.
• All other uncredited images are from research done at the EWI faculty.